A wire X of length 50 ; cm carrying a current | vedclass

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(A) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $r$ is given by $f = \frac{\mu_0 I_

Vedclass · Accepted Answer

$1.2 	imes 10^{-5} \; N$ directed towards wire $Y$.

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(A) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi r}$.Here,$I_1 = 2 \; A$,$I_2 = 3 \; A$,$r = 5 \; cm = 0.05 \; m$,and the length of wire $X$ is $\ell = 50 \; cm = 0.5 \; m$.The total force on wire $X$ is $F = f 	imes \ell = \frac{\mu_0 I_1 I_2 \ell}{2 \pi r}$.Substituting the values:$F = \frac{(4 \pi 	imes 10^{-7} \; T \cdot m/A) 	imes (2 \; A) 	imes (3 \; A) 	imes (0.5 \; m)}{2 \pi 	imes (0.05 \; m)}$$F = \frac{2 	imes 10^{-7} 	imes 6 	imes 0.5}{0.05} = \frac{6 	imes 10^{-7}}{0.05} = 120 	imes 10^{-7} = 1.2 	imes 10^{-5} \; N$.Since the currents are in the same direction,the force is attractive,meaning wire $X$ is pulled towards wire $Y$.

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